Question:

Step 1: The slope of the curve is given by $\frac{dy}{dx} = \frac{x^2 - 2y}{x}$. This can be rewritten as a linear first-order differential equation: $$ \frac{dy}{dx} = x - \frac{2y}{x} $$ $$ \frac{dy}{dx} + \frac{2}{x}y = x $$ This is in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \frac{2}{x}$ and $Q(x) = x$. The integrating factor (IF) is $e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = e^{\ln(x^2)} = x^2$. Multiplying the differential equation by the integrating factor: $$ x^2 \frac{dy}{dx} + 2xy = x^3 $$ The left side is the derivative of the product $x^2y$: $$ \frac{d}{dx}(x^2y) = x^3 $$ Integrating both sides with respect to $x$: $$ \int \frac{d}{dx}(x^2y) dx = \int x^3 dx $$ $$ x^2y = \frac{x^4}{4} + C $$ Step 2: The curve passes through the point $(1, -2)$. Substitute $x=1$ and $y=-2$ into the general solution to find the constant $C$: $$ (1)^2(-2) = \frac{(1)^4}{4} + C $$ $$ -2 = \frac{1}{4} + C $$ $$ C = -2 - \frac{1}{4} = -\frac{8}{4} - \frac{1}{4} = -\frac{9}{4} $$ Thus, the particular solution is: $$ x^2y = \frac{x^4}{4} - \frac{9}{4} $$ Solving for $y$: $$ y = \frac{x^4}{4x^2} - \frac{9}{4x^2} $$ $$ y = \frac{x^2}{4} - \frac{9}{4x^2} $$ Step 3: To find the ordinate of the point on the curve with abscissa $x = -1$, substitute $x=-1$ into the particular solution: $$ y(-1) = \frac{(-1)^2}{4} - \frac{9}{4(-1)^2} $$ $$ y(-1) = \frac{1}{4} - \frac{9}{4(1)} $$ $$ y(-1) = \frac{1}{4} - \frac{9}{4} $$ $$ y(-1) = -\frac{8}{4} $$ $$ y(-1) = -2 $$